2 Answers
Lokesh Verma
·2010-02-21 04:09:21
\\\frac{1}{1+\frac{A_2}{A_1}}+\frac{1}{1+\frac{A_4}{A_3}}-\frac{2}{1+\frac{A_3}{A_2}} \\\frac{1}{1+\frac{n-2}{2}}+\frac{1}{1+\frac{n-4}{4}}-\frac{2}{1+\frac{n-3}{3}} \\\frac{1}{\frac{n}{2}}+\frac{1}{\frac{n}{4}}-\frac{2}{\frac{n}{3}}=0